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    Chapter 2. Special Relativity

    Notes: • Some material presented in this chapter is taken “The Feynman Lectures on

    Physics, Vol. I” by R. P. Feynman, R. B. Leighton, and M. Sands, Chap. 15 (1963, Addison-Wesley).

    2.1 The Ether and the Michelson-Morley Experiment As we saw at the end of Chapter 1, the prediction and experimental verification of the propagation of electromagnetic waves in space led physicists to the conclusion that there should exist a medium, the ether, permeating all of space on which these waves travelled. Of course, the wave equation we derived from Maxwell’s equations (see equation (1.24) in Chapter 1) does not at all require that. It does, in fact, imply that such waves could propagate in vacuum. But this notion of wave propagation without a medium was so counterintuitive to physicists at the time that they elected to postulate the existence of the ether. On the other hand, this hypothesis had the advantage of being testable through experiments. This is exactly what American physicists Albert Michelson (1852-1931) and Edward Morley (1838-1923) sought out to do in a famous experiment (i.e., the Michelson- Morley experiment) in 1887. To do so they used an apparatus (now called a Michelson interferometer) as shown in the schematic of Figure 1. In a nutshell, the experiment consists of sending a light signal (from Source A in the figure) and splitting it over two mutually orthogonal paths (at the plate B), each propagating through a distance L to a mirror (mirrors C and E in the up-down and left-right directions, respectively) where they are reflected back and recombined (“below” the plate B).



    B B0

    C C0

    D D0

    E E0

    F F 0

    u L



    Waves in phase

    Waves out of phase

    Fig. 15-2. Schematic diagram of the Michelson-Morley experiment.

    The plate B splits an oncoming beam of light, and the two resulting beams continue in mutually perpendicular directions to the mirrors, where they are reflected back to B. On arriving back at B, the two beams are recombined as two superposed beams, D and F . If the time taken for the light to go from B to E and back is the same as the time from B to C and back, the emerging beams D and F will be in phase and will reinforce each other, but if the two times di�er slightly, the beams will be slightly out of phase and interference will result. If the apparatus is “at rest” in the ether, the times should be precisely equal, but if it is moving toward the right with a velocity u, there should be a di�erence in the times. Let us see why.

    First, let us calculate the time required for the light to go from B to E and back. Let us say that the time for light to go from plate B to mirror E is t1, and the time for the return is t2. Now, while the light is on its way from B to the mirror, the apparatus moves a distance ut1, so the light must traverse a distance L + ut1, at the speed c. We can also express this distance as ct1, so we have

    ct1 = L + ut1, or t1 = L/(c ≠ u).

    (This result is also obvious from the point of view that the velocity of light relative to the apparatus is c ≠ u, so the time is the length L divided by c ≠ u.) In a like manner, the time t2 can be calculated. During this time the plate B advances a


    Figure 1 - Schematic diagram of the Michelson- Morley experiment (from The Feynman Lectures on Physics, Vol. I).

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    Michelson and Morley surmised that if the earth is moving relative to the ether at a speed u in the B-E direction, for example, then the electromagnetic wave from the light source travelling along that direction should make its round trip from plate B to mirror E and back to plate B in a different amount of time than the wave propagating along the B-C-B path. More precisely, since the apparatus is moving in the B-E direction the corresponding velocity of the light wave relative to the interferometer should be (according the Newtonian mechanics) equal to c − u . It follows that the time needed for the light wave to go from B to E is

    t1 = L

    c − u . (2.1)

    Once the wave reflects on mirror E, its velocity relative to the apparatus becomes equal to c + u and the time needed to travel from E to B is

    t2 = L

    c + u . (2.2)

    We then find that the time necessary for the round trip is

    tE = t1 + t2

    = 2L c 1− u2 c2( ) .


    For the light wave going from plate B to mirror C and back the speed relative to the interferometer is simply c , since its velocity is perpendicular to that of the apparatus. The distance the light goes through when going from B to C is given by

    d3 = ut3( )2 + L2 , (2.4) where t3 is the time needed for the wave to travel that distance. But since we also have t3 = d3 c , we can write ct3( )2 = ut3( )2 + L2, (2.5) or

    t3 = L c

    1− u2 c2 . (2.6)

    It follows that the round trip B-C-B is

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    tC = 2L c 1− u2 c2

    . (2.7)

    According to the ether hypothesis, the time difference between the two paths should have been

    Δt = tC − tE

    = 2L c 1− u2 c2

    1−1 1− u2 c2( ) ≠ 0.


    This non-zero time difference Δt predicted that Michelson and Morley would observe that, when recombined, the two light waves should not be in phase with each other. That is, if when they left the plate B at the initial time t0 the waves had the same amplitude E t0( ) = Acos φ0( ), (2.9) then, when recombined, the total amplitude of the signal should had been

    ET t( ) = Acos ωtC +φ0( ) + Acos ωtE +φ0( ) = A cos ωtC +φ0( ) + cos ωtC −ωΔt +φ0( )⎡⎣ ⎤⎦ = 2Acos ωΔt

    2 ⎛ ⎝⎜

    ⎞ ⎠⎟ cos ωtC +φ0 −

    ωΔt 2

    ⎛ ⎝⎜

    ⎞ ⎠⎟ ,


    where we used the identity cos θ1( )cos θ2( ) = cos θ1 −θ2( ) + cos θ1 +θ2( )⎡⎣ ⎤⎦ 2 . However, the signal Michelson and Morley detected corresponded to ET t( ) = 2Acos ωtC +φ0( ), (2.11) which implied that Δt = 0 ! In other words, the time taken by the light waves to travel their respective paths was the same, just as if the speed of light was the same in both orientations. This result was directly at odds with the idea that electromagnetic waves propagated on the hypothetical ether. It interesting to note the observation of Lorentz who suggested that the result could be explained if bodies (like the B-E leg of the Michelson interferometer) contracted according to LE = L 1− u

    2 c2 , (2.12)

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    in the direction of their velocity u . It could indeed verified from equation (2.8) that Δt = 0 in that case even with the existence of the hypothetical ether. Although this suggestion was too artificial to be satisfying to physicists at the time, we will soon see that it incorporated part of the solution to the problem exposed by the Michelson-Morley experiment.

    2.2 The Invariance of Maxwell’s Equations Let us now revisit the earlier statement made in Section 1.2.2 of Chapter 1 that Maxwell’s equations were not invariant under a Galilean transformation. More precisely, let us start with the wave equation we derived from Maxwell’s equations in Chapter 1 (i.e., equation (1.24))

    ∂2Ez ∂x2

    − 1 c2

    ∂2Ez ∂t 2

    = 0 (2.13)

    and let us see how it transforms when going to a reference frame moving at a relative speed u in the positive x-direction such that

    ′t = t ′x = x − ut ′y = y ′z = z.


    We now use the chain rule to write

    ∂ ∂x

    = ∂ ′x ∂x

    ∂ ∂ ′x

    + ∂ ′t ∂x

    ∂ ∂ ′t

    = ∂ ∂ ′x


    ∂x2 = ∂


    ∂ ′x 2



    ∂ ∂t

    = ∂ ′t ∂t

    ∂ ∂ ′t

    + ∂ ′x ∂t

    ∂ ∂ ′x

    = ∂ ∂ ′t

    − u ∂ ∂ ′x


    ∂t 2 = ∂

    ∂ ′t − u ∂

    ∂ ′x ⎛ ⎝⎜

    ⎞ ⎠⎟

    ∂ ∂ ′t

    − u ∂ ∂ ′x

    ⎛ ⎝⎜

    ⎞ ⎠⎟

    = ∂ 2

    ∂ ′t 2 − 2u ∂


    ∂ ′x ∂ ′t + u2 ∂


    ∂ ′x 2 .


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    Inserting the last of equations (2.15) and (2.16) into equation (2.13) yields1

    1− u 2

    c2 ⎛ ⎝⎜

    ⎞ ⎠⎟ ∂2Ez ∂ ′x 2

    − 1 c2

    ∂2Ez ∂ ′t 2

    + 2u c2

    ∂2Ez ∂ ′x ∂ ′t

    = 0. (2.17)

    Because physicists believed then, as we still do now, that the laws of physics should be the same in any and all inertial frames, it follows that the fact that equation (2.17) does not “look” the same as equation (2.13) was a major problem. By “looking” the same we mean that if we replace x and t with ′x and ′t in equation (2.13) we do not recover equation (2.17), and vice-versa. Indeed, we can see that the two equations are similar only when u≪ c . It was once again Lorentz who pointed out a fact that would eventually be central to resolving this problem. More precisely, he discovered that if one instead used the following coordinate transformation between the two inertial frames

    ′t = t − xu c2

    1− u2 c2

    ′x = x − ut 1− u2 c2

    ′y =


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